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Proving that the intersection of two sets is a subset of Q?

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data
    "Let ##U={p+r\sqrt{2}:p,r∈ℚ}## and let ##V={a+b\sqrt{7}:a,b∈ℚ}##. Show that ##ℚ⊆U∩V##. Then show that ##U∩V⊆ℚ## and conclude that ##U∩V=ℚ##."

    2. Relevant equations


    3. The attempt at a solution
    I only knew how to prove the first part. That is:

    (1)
    "Suppose that ##ℚ⊄U∩V##. This implies that ##∃x∈ℚ## such that ##x∉U## and ##x∉V##. But since ##ℚ⊂U## and ##ℚ⊂V##, it follows that there does not exist an ##x∈ℚ⊂U## such that ##x∉U##. Similarly, there exist no ##x∈ℚ⊂V## such that ##x∉V##."

    I'm also ambivalent about saying that ##ℚ⊂U## when I haven't proven yet that ##∃q## such that ##p+q\sqrt{2}∉ℚ##. I'm thinking about just changing it to ##ℚ⊆U##.

    (2)
    ##{p+r\sqrt{2}:p,r∈ℚ}∩{a+b\sqrt{7}:a,b∈ℚ}##

    And here is where I got stuck; proving that if ##r≠0##, then ##p+r\sqrt{2}∉V##. Similarly, I was unable to prove that ##a+b\sqrt{7}∉U##. Basically, I'm trying to prove that every element of an intersection whose elements I am unable to properly identify. I'm thinking of just doing it one piece at a time:

    "Suppose ##p+r\sqrt{2}∈V##. Then we can say that ##p+r\sqrt{2}=a+b\sqrt{7}##. With algebra, I get: ##\sqrt{2}=\frac{1}{r}(s+b\sqrt{7})## where ##s=a-p##. Squaring both sides gives me ##2=\frac{1}{r}(s^2+2bs\sqrt{7}+7b^2)##. However, contradicts the fact that ##2## is a prime number, since it can only be written as the product of two integers: itself and ##1##."

    That's only proving that any element of ##U## with ##r≠0## is an element of ##V##, and I don't even think I'd be done, since I cannot rule out the possibility that ##s+b\sqrt{7}=1## and ##\frac{1}{r}=2##; or the possibility that ##\frac{1}{r}=1## and ##s+b\sqrt{7}=\sqrt{2}##.

    I was given a hint that I needed to use the fact that ##\sqrt{14}∉ℚ## for the second part...
     
  2. jcsd
  3. Mar 8, 2017 #2

    Mark44

    Staff: Mentor

    Think about what the elements of ##U \cap V## can be. It's also helpful to think about things from a vector space perspective. If you have a nonzero element q of ##\mathbb{Q}## (a rational number), any other rational number is some (rational) multiple of q. In linear algebra terms, q spans ##\mathbb{Q}##. Set U can be thought of as all linear combinations of elements from ##\mathbb{Q}## and ##\sqrt 2##. With set U, you get all possible rational numbers, plus something else -- all rational multiples of ##\sqrt 2##. Set V is something like this, with all possible rational numbers, plus all rational multiples of ##\sqrt 7##.

    So I'm really thinking of U as being spanned by two "vectors" -- 1 and ##\sqrt 2##, and V being spanned by a different set of vectors -- 1 and ##\sqrt 7##. What do the two sets have in common?
     
  4. Mar 8, 2017 #3

    vela

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    That should be ##x∉U## OR ##x∉V##.

    In any case, your proof seems unnecessarily complicated. If ##x \in \mathbb{Q}##, what can you say about ##x + 0\sqrt{2}## and ##x+0\sqrt{7}##?
     
  5. Mar 9, 2017 #4
    They have ##1## as that common element. So how would I express the fact that the sets are in different planes, except at the ##ℚ## axis (so to speak) where they intersect?

    They're both elements of ##U## and ##V##.
     
  6. Mar 9, 2017 #5

    vela

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    If ##x \in U\cap V##, you have ##x = p+r\sqrt{2} = a+b\sqrt{7}##. Rearrange terms to get ##p-a = b\sqrt{7}-r\sqrt{2}## and then square both sides. See where that takes you.
     
  7. Mar 9, 2017 #6
    Let's see...

    ##p^2-2pa+a^2=7b^2-2br\sqrt{14}+2r^2##
    ##-\frac{p^2-2pa+a^2-7b^2-2r^2}{2br}=\sqrt{14}##

    So I get the contradiction that ##\sqrt{14}## is supposed to be irrational? Or is it to early to say that?
     
    Last edited: Mar 9, 2017
  8. Mar 9, 2017 #7

    Mark44

    Staff: Mentor

    ???
    ##\sqrt{14}## is irrational. The left side, ##p^2 - 2pa + a^2## is rational, as are ##7b^2## and ##2r^2##. Can you have a rational number on one side of an equation, and an irrational one on the other side?
     
  9. Mar 9, 2017 #8
    No; and that's where I would get the contradiction from. Okay, I think I got my answer. Thanks.
     
    Last edited: Mar 9, 2017
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